Core

http://www.examsolutions.net/a-level-maths-papers/Edexcel/Core-Maths/Core-Maths-C3/2007-January/paper.php#Q6

how do i draw a graph of

$ln1-x?$

i can drawn $ln1+x\ or\ lnx-1\ or\ ln2x\ or\ lnx^2$

but they never taught me how to draw a graph of ln something minus x how do i do it?

It will just be a few simple transformations (reflections, translations) so the graph will still look similar.
You may need to just re write it using latex because I can't really understand what you have put.

You realise ln1 is just 0 so in fact you're just drawing -x which is just a straight line through the origin with a negative gradient

You realise ln1 is just 0 so in fact you're just drawing -x which is just a straight line through the origin with a negative gradient

you want some brackets? i guess i should probably put them in, check back again after you get the notification

yes but surely this is the transformation of and thus i'd just move the graph right 1 unit?

but yea i want to reflect it in the y-axis if it's -x

He (obviously) means $\ln (1-x)$.

@OP. Denote $f(x) = \ln x$ then $g(x) = f(-x) = \ln (-x)$ is a reflection in the $y$-axis of $f(x)$. Draw this.

Now $h(x) = g(x+1) = \ln(-x +1) = \ln(1-x)$ which is a translation to the left by one unit of $g(x)$. Can you now draw this?

This is just f(-x) as the a part is 0 because ln1=0
Edit: just saw Zacken's post. Didn't realise there were meant to be brackets.

so 1 unit to the left of the graph $y=\ln x$ reflected in the y-axis?

Yes.

awesome

so what about the flipside

$y=2-\dfrac{1}{2} e^x$

what about one of these?

You work through it for yourself systematically first before opening the spoiler below, it's all very procedural.

i drew the graph like this.
$y=e^x\ y=\frac{1}{2}e^x[br]y=-\frac{1}{2}e^x\ y=2-\frac{1}{2}e^x$

i wasn't sure whether it went above the x-axis or not :/

1. Your attempts to use LaTeX aren't helping, could you please stick to/at least provide ascii typed mathematical expressions of the form y = e^x, y = (1/2) e^x, etc... so that we may understand what you are saying.

2. You should know basic properties of $\exp$. It has an asymptote at the $x$-axis, so $y=0$. The first transformation, halving the function "changes" the asymptote to $y = \frac{0}{2} = 0$. The second transformation, reflecting the curve "changes the asymptote to $y = -\frac{0}{2} = 0$. The third transformation shifts the asymptote 2 units upwards, so changes it to $y = -\frac{0}{2} + 2 = 2$. So your graph is correct, indeed it is asymptotic to the line $y = 2$ as $x \to -\infty$ and then diverges off to negative infinity as $x \to \infty$. Furthermore, it has a y-intercept of $(0, 1)$ and an $x$-intercept at $(\ln 2, 0)$.

So y-intercept means x=0

so if i stick x=0 into $y=2-\frac{1}{2} e^x$ e^x becomes 1
1x0.5=0.5
so 2-0.5=1.5 which isn't 1 or did i go wrong somewhere?

and it touches the x axis at y=0
so when y=0
$0=2-\frac{1}{2} e^x$

$2=\frac{1}{2} e^x$

$4=e^x$

$ln4=x$

where did it all go wrong?

Yeah, that's all fine. I thought the function was $2 - e^x$ whilst writing my reply up, hence the discrepancy.